Question

A bullet of mass 10 g moving horizontally with speed of 500 m/s passes through a
block wood of mass 1 kg, initially at rest on frictionless surface. The bullet comes
out of the block with a speed of 200 m/s. The final speed of the block is
A) 500 m/s B) 300 m/s C) 200 m/s D) 3 m/s

Answer 1

Answer:

The correct answer will be 3 m/s means option D

Explanation:

According to the problem the mass of the bullet , m1 = 10  g = 0.01 kg

The initial velocity of the bullet is, u1 = 500 m/s

The mass of the block is m2 = 1 kg

The initial velocity of the block u2 = 0

The final velocity of the block is v= 200 m/s

Le the final velocity of the block is v1

Therefore,

m1u1 + m2u2 = m1v +m2v1

=> 0.01 x 500 + 1 x0 = 0.01 x 200 + 1 x v1

=> v1 = 3 m/s

Answer 2

Answer:

Correct Option: D) 3 m/s

Note: This question had been asked in JEE Mains 2016, 2013 and in CUCET 2016.

Explanation:

m1 = 10 g = 0.01 kg

u1 = 500 m/s

v1 = 200 m/s

m2 = 1 kg

u2 = 0 m/s

v2 = ?

In this case there is Elastic Collision, we know that in elastic collision, momentum is conserved. So,

m1u1 + m2u2 = m1v1 + m2v2

(0.01 kg X 500 m/s) + (1 X 0) = (0.01 kg X 200 m/s) + (1 X v2)

5 = 2 + v2

v2 = 3 m/s